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I'm trying to find the derivative of $f=\arcsin\left(\frac{2x}{1+x^2}\right)$. I think I'm mistaken and perhaps using the chain rule incorrectly.

Let $g(x) = \frac{2x}{1+x^2}$ and let $h(x) = \arcsin x$

According to the chain rule -

$$f'(x) = \frac{1}{\sqrt{1-\frac{2x}{1+x^2}}}⋅((2⋅(1+x^2 )-2x⋅2x)/(1+x^2 )^2 ) = \cdots \frac{-2(x^4-1)}{x-1}$$

Is this a correct usage of the chain rule?

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    The derivative of $\arcsin$ is $\dfrac1{\sqrt{1-x^2}}$, not $\dfrac1{\sqrt{1-x}}$. Other than that, the first passage seems ok. – 2017-01-31
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    Can one begin by establishing the trigonometric identity $$ \arcsin\left( \frac{2x}{1+x^2} \right) = 2\arctan x \text{ ?} $$ Does this hold only if $-1\le x\le 1$ and do we need absolute-value signs or the like if $x>1$ or $x<-1 \text{ ?}\qquad$ – 2017-01-31

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\begin{align} \frac d {dx} \arcsin\left( \frac{2x}{1+x^2} \right) & = \frac 1 {\sqrt{1-\left( \frac{2x}{1+x^2} \right)^2}} \cdot \frac d {dx} \frac{2x}{1+x^2} & & \text{by the chain rule} \\[10pt] & = \frac 1 {\sqrt{1-\left( \frac{2x}{1+x^2} \right)^2}} \cdot \frac{2(1-x^2)}{(1+x^2)^2} & & \text{by the quotient rule} \\[10pt] & = \frac 1 {\sqrt{\frac{(1+x^2)^2 - (2x)^2}{(1+x^2)^2} }} \cdot \frac{2(1-x^2)}{(1+x^2)^2} \\[10pt] & = \frac{1+x^2}{\sqrt{(1+x^2)^2 - (2x)^2}} \cdot \frac{2(1-x^2)}{(1+x^2)^2} \\[10pt] \end{align}

Now notice that $$ (1+x^2)^2 - (2x)^2 = (1+2x^2+x^4) - 4x^2 = 1-2x^2 + x^4 = (1-x^2)^2 $$ so we get \begin{align} \sqrt{(1+x^2)^2 - (2x)^2} = |1-x^2|. \end{align} If $-1\le x\le1$ then this is $1-x^2$ and so by cancellations from the numerator and denominator the derivative simplifies to $$ \frac 2 {1+x^2} \qquad \left( = 2 \frac d {dx} \arctan x \right). $$ However, since arcsine is not differentiable at $\pm1,$ the chain rule does not tell us that the composite function is differentiable at points where $2x/(1+x^2)\in\{\pm1\}.$ Those two points must therefore be examined separately.

If $x>1$ or $x<-1$ then $|1-x^2| = -(1-x^2)$ and the answer we get will be multiplied by $-1$.

Notice that $2x/(1+x^2)=\pm1$ when $x=\pm1$ and has absolute extreme values at those points. Thus $2x/(1+x^2)$ is never outside the domain of the arcsine function.

So the derivative is $\dfrac{-2}{1+x^2}$ when $x>1$ or $x<-1$.

When $x=\pm1$ then $2x/(1+x^2)=\pm1$ and one can find the two one-sided derivatives at those points. Lo and behold, they are not equal, so the function is not differentiable at those two points. $$ \frac d {dx} \arcsin\left( \frac{2x}{1+x^2}\right) = \begin{cases} \dfrac 2 {1+x^2} & \text{if } -11 \text{ or } x<-1. \end{cases} $$

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    The derivative is not $\frac{2}{1+x^2}$. – 2017-01-31
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    @egreg : Did you read the last two sentences? – 2017-01-31
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    Yes, but this could be very misleading to a beginner. That's why I decided to write a proper answer: this one might be good if it were accurate about the absolute value, two are grossly wrong, another one gets the result from nowhere. – 2017-01-31
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Your mistake here.

$$\sin^{-1} \frac{2x}{1+x^2} = \frac1{\sqrt{1-\left(\frac{2x}{1+x^2}\right)^2}}.$$

Its not $\sin^{-1} \dfrac{2x}{1+x^2} = \dfrac1{\sqrt{1-\frac{2x}{1+x^2}}}$.

After this your quotient rule is fine.

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    Actually I always use this site on my mobile. So my browser not update any answer or comment until I refresh. And in the mean time I am busy in typing answer. – 2017-01-31
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    @amWhy if you want me to delete my answer. Its ok. – 2017-01-31
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With the chain rule: $$f'(x)=\frac{1}{\sqrt{1-\left(\frac{2x}{1+x^2}\right)^2}}\left(\frac{2(1+x^2)-2x(2x)}{(1+x^2)^2}\right)=\frac{1+x^2}{\sqrt{(1+x^2)^2-4x^2}}\left(\frac{2-2x^2}{(1+x^2)^2}\right)$$ $$=\frac{2-2x^2}{1+x^2}\frac{1}{|1-x^2|}=\frac{2}{1+x^2}\frac{1-x^2}{|1-x^2|}$$

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    You are right, I do not consider the absolute value when I did the root, so now I change my answer. Thanks – 2017-02-01
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Since it's easy to graph the function with some software (also online) it should be clear that the derivative cannot be everywhere positive and the graph also suggests the function is not differentiable at $-1$ and $1$.

enter image description here

Set $g(x)=\dfrac{2x}{1+x^2}$; then $$ f(x)=\arcsin\dfrac{2x}{1+x^2}=\arcsin(g(x)) $$ and, by the chain rule, $$ f'(x)=\frac{1}{\sqrt{1-(g(x))^2}}g'(x) $$ Now divide et impera:

  1. $\displaystyle 1-(g(x))^2=\frac{(1+x^2)^2-4x^2}{(1+x^2)^2} =\frac{(1-x^2)^2}{(1+x^2)^2}$

  2. $\displaystyle \frac{1}{\sqrt{1-(g(x))^2}}=\frac{1+x^2}{|1-x^2|}$

  3. $\displaystyle g'(x)=2\frac{1+x^2-x\cdot2x}{(1+x^2)^2}$

Therefore $$ f'(x)=\frac{1+x^2}{|1-x^2|}\frac{2(1-x^2)}{(1+x^2)^2}= \frac{2}{1+x^2}\frac{1-x^2}{|1-x^2|} $$

The derivative does exist at $-1$ and $1$, as it's easy to check with the limits from the left and from the right.

By the way, this shows that $$ f(x)=\begin{cases} c_--2\arctan x & \text{if $x<-1$} \\ c_0+2\arctan x & \text{if $-1\le x\le 1$}\\ c_+-2\arctan x & \text{if $x>1$} \end{cases} $$ Since $$ \lim_{x\to-\infty}f(x)=0,\qquad f(0)=0,\qquad \lim_{x\to\infty}f(x)=0 $$ we can conclude that $$ c_-=-\pi,\qquad c_0=0,\qquad c_+=\pi $$

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    First of all thank you for the fix. In your first calculation of $1-g(x)^2$, isn't it $\frac{(x^2-1)^2}{(1+x^2)^2}$? Actually now I realize that since it is $()^2$ opposite $+-$ does not matter. Moreover, which software do you use to graph functions? – 2017-01-31
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    @S.Peter $(1-x^2)^2=(x^2-1)^2$. The software is Grapher, comes with Mac OS X. – 2017-01-31
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    Sorry I bring this up now but could you please show also how to find the limits from both sides of $x=1, -1$? For example, how can I find $\lim_{x\to1^+}\frac{\arcsin \frac{2x}{1+x^2} - \pi/2}{x-1}$ – 2017-02-04
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    @S.Peter Use l'Hôpital, of course: you already have the derivative, haven't you? `;-)` – 2017-02-04
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If $\arctan x=y,-\dfrac\pi2\le y\le\dfrac\pi2, x=\tan y$

$\implies\dfrac{2x}{1+x^2}=\sin2y$

$\implies\arcsin\dfrac{2x}{1+x^2}=\begin{cases} 2\arctan x &\mbox{if }-\dfrac\pi2\le2y\le\dfrac\pi2\iff-1\le x\le1 \\\pi-2\arctan x & \mbox{if } 2y>\dfrac\pi2\iff x>1 \\-\pi-2\arctan x & \mbox{if } 2y<-\dfrac\pi2\iff x<-1\end{cases}$

Now $\dfrac{d(\arctan x)}{dx}=?$

Actually, $\sqrt{(1+x^2)^2-4x^2}=\sqrt{(1-x^2)^2}=|1-x^2|= \begin{cases} 1-x^2 &\mbox{if }1-x^2\ge0\iff-1\le x\le1\\ x^2-1& \mbox{if }x^2>1\iff x>1\text{ or }x<-1\end{cases}$